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3 years ago

Heather is going to graph y= -3x^2 +3. How is the parent graph transformed

1 answer:

6 0

Well compared to the parent function x^2, the graph of y=-3x^2+3 is flipped upside down (negative sign), steeper, and is up 3 units from the origin.

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The point (4,7) is on the graph of y=x^2+c. what is the value of c?

slamgirl [31]

Point is (4,7)
So y = 7 = 4^2 + c
c = 7 - 16
c = -9

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3 years ago

4x^2+4y^2+20x-16y+37=0 standard form

zhenek [66]

The answer in standard form is (x+5/2)^2+(y-2)^2=1

5 0

3 years ago

there are between 24 to 40 students in a class. the ratio of boys to girls is 4:7. how many students are in the class.

ZanzabumX [31]

Answer:

8 students

Step-by-step explanation:

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2 years ago

Read 2 more answers

Explain why a graph that fails the vertical-line test does not represent a function

Anon25 [30]

For any given y value, there can only be one x value. Let's consider a graph with a line that curves and intersects x=2 twice. This would mean that if you plug 2 into the function you would get two answers. However, this is not possible as plugging in any value to x will result in only one answer. For example, if y= 4x + 3, no matter how many times I plug a number in there, I'm always going to get the same y value, and this is true for all real functions.

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3 years ago

Given: AB || DE , AD bisects BE. Prove: ABC = DEC using the ASA postulate.

Ket [755]

Answer:

As per ASA postulate, the two triangles are congruent.

Step-by-step explanation:

We are given two triangles:

$\triangle ABC$ and $\triangle DEC$ .

AD bisects BE.

AB || DE.

Let us have a look at two properties.

1. When two lines are parallel and a line intersects both of them, then alternate angles are equal.

i.e. AB || ED and $\angle B$ and $\angle E$ are alternate angles $\Rightarrow$ $\angle B = \angle E$ .

2. When two lines are cutting each other, angles formed at the crossing of two, are known as Vertically opposite angles and they are are equal.

$\Rightarrow \angle ACB = \angle DCE$

Also, it is given that AD bisects BE.

i.e. EC = CB

1. $\angle B = \angle E$

2. EC = CB

3. $\angle ACB = \angle DCE$

So, we can in see that in $\triangle ABC$ and $\triangle DEC$ , two angles are equal and side between them is also equal to each other.

Hence, proved that $\triangle ABC$ $\cong$ $\triangle DEC$ .

8 0

2 years ago